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I'm trying to generalize this question to arbitrary $n$ dimensions, i.e. to show that $C = \{x\in \mathbb R^n_{++} : \prod_i^nx_i \ge 1 \}$ is convex.

At first I thought maybe through induction, but I reached a dead end there. Any suggestions?

Maverick Meerkat
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2 Answers2

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The set can be expressed as $$\left\{ x \in \mathbb{R}_{++}^n : \sum_{i=1}^n \log x_i \geq 0 \right\},$$ which is convex as $g(x)=\sum_{i=1}^n \log x_i$ is a concave function.

LinAlg
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  • can you elaborate why a set is convex if it's an inequality involving a function of concave? – Maverick Meerkat May 19 '20 at 13:43
  • This is the good answer. – Didier May 19 '20 at 13:43
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    @DavidRefaeli suppose $x$ and $y$ are in $C$, then $g(x)\geq 0$ and $g(y)\geq 0$, so $g(\lambda x + (1-\lambda)y) \geq \lambda g(x) + (1-\lambda) g(y) \geq 0$ by concavity of $g$, so the midpoint is also in $C$ (for any $0\leq \lambda \leq 1$). – LinAlg May 19 '20 at 13:47
  • yeah also by the fact that -concave = convex. And a level set of convex function (i.e. convex < c) is convex. – Maverick Meerkat Jun 13 '21 at 16:19
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By weighted AM GM, for any $(x_i), (y_i) \in C$ and $t \in (0,1)$, we have

$$\prod_{i=1}^n (tx_i + (1-t)y_i) \ge \prod_{i=1}^n x_i^t y_i^{1-t} = \left( \prod_{i=1}^n x_i\right)^t \left(\prod_{i=1}^n y_i\right)^{1-t} \ge 1^t 1^{1-t} = 1$$

achille hui
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